Question

Consider samples of phosphine (PH₃), water (H₂O), hydrogen sulfide (H₂S), and hydrogen fluoride (HF), each with a mass of 119 g. Rank the compounds from the least to the greatest number of hydrogen atoms contained in the samples.

Short answer

The compounds from the least to the greatestnumber of hydrogen atoms contained in the samples, H₂S < PH₃< H₂O.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

Molar mass =$\frac{Weighting}{numberofmoles}$

Determination of molar mass of PH3

Here, the molar mass of PH₃:

$30.97\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.+1.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol\times 3=34.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$}\right.$

Then the amount of hydrogen atoms:

$119g\times \frac{1mol}{34.00g}\times \frac{3molH}{1molP{H}_3}\times \frac{6.02\times {10}^{23}atoms}{1mol}=6.32\times {10}^{24}atoms$

Determination of molar mass of H2O

Here, the molar mass of H₂O:

$1.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\times 2+16.00\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.=18.02\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Then the amount of hydrogen atoms:

$119g\times \frac{2mol}{18.02g}\times \frac{2molH}{1mol{H}_{2}O}\times \frac{6.02\times {10}^{23}atoms}{1mol}=7.95\times {10}^{24}atoms$

Determination of molar mass of H2S

Here, the molar mass of H₂S:

$1.01\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\times 2+32.07\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.=34.09\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Then the amount of hydrogen atoms:

$119g\times \frac{1mol}{34.09g}\times \frac{2molH}{1mol{H}_{2}O}\times \frac{6.02\times {10}^{23}atoms}{1mol}=4.20\times {10}^{24}atoms$

Therefore, H₂S < PH₃ < H₂O.